Subjects algebra

Function Ranges 3A4953

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1. **State the problem:** We have three functions: $$f(x) = - \frac{6}{11} \left(\frac{11}{2}\right)^x, \quad g(x) = \frac{6}{11} \left(\frac{11}{2}\right)^{-x}, \quad h(x) = - \frac{6}{11} \left(\frac{11}{2}\right)^{-x}$$ We need to determine which of the given statements about their domains and ranges is true. 2. **Recall domain and range rules for exponential functions:** - The domain of any exponential function $a^x$ where $a>0$ and $a \neq 1$ is all real numbers $(-\infty, \infty)$. - The range depends on the coefficient and the base. - For $a^x$, range is $(0, \infty)$ if coefficient is positive, and $(-\infty, 0)$ if coefficient is negative. - For $a^{-x} = \frac{1}{a^x}$, the domain remains all real numbers, and range is similar to $a^x$ but flipped in terms of $x$. 3. **Analyze each function:** - For $f(x) = - \frac{6}{11} \left(\frac{11}{2}\right)^x$: - Domain: all real numbers. - Since coefficient is negative, range is $(-\infty, 0)$. - For $g(x) = \frac{6}{11} \left(\frac{11}{2}\right)^{-x} = \frac{6}{11} \cdot \frac{1}{\left(\frac{11}{2}\right)^x}$: - Domain: all real numbers. - Coefficient positive, so range is $(0, \infty)$. - For $h(x) = - \frac{6}{11} \left(\frac{11}{2}\right)^{-x} = - \frac{6}{11} \cdot \frac{1}{\left(\frac{11}{2}\right)^x}$: - Domain: all real numbers. - Coefficient negative, so range is $(-\infty, 0)$. 4. **Check each statement:** - "The range of $h(x)$ is $y > 0$" is false because $h(x)$ has negative coefficient, so range is $y < 0$. - "The domain of $g(x)$ is $y > 0$" is false because domain refers to $x$ values, not $y$ values. Domain of $g(x)$ is all real numbers. - "The ranges of $f(x)$ and $h(x)$ are different from the range of $g(x)$" is true because $f(x)$ and $h(x)$ have range $(-\infty, 0)$, while $g(x)$ has range $(0, \infty)$. - "The domains of $f(x)$ and $g(x)$ are different from the domain of $h(x)$" is false because all have domain all real numbers. **Final answer:** The true statement is: **The ranges of $f(x)$ and $h(x)$ are different from the range of $g(x)$.**